REVIEW OF CALCULATOR FUNCTIONS
FOR THE TEXAS INSTRUMENTS BA II PLUS@
Samuel Broverman, University of Toronto
This note presents a review of calculator financial functions for the Texas
Instruments BA II PLUS calculator. This note, including a number of the
examples used as illustrations, is reprinted with permission from the 3rd
edition of the book Mathematics of Investment and Credit, by S.
Broverman. Also, several examples from SOA/CAS math of finance
exams (old Course 2) will be presented illustrating the use of the
calculator.
A detailed guidebook for the operation of and functions available on the
BA II PLUS can be found at the following internet site:
http://education.ti.com/us/globallguides.html#finance. It will be assumed
that you have available and have reviewed the appropriate guide book for
the calculator that you are using.
Financial functions will be reviewed in the order that the related concepts
are covered in Chapters 1 to 8 of Mathematics of Investment and Credit.
Some numerical values will be rounded off to fewer decimals than are
actually displayed in the calculator display.
It will be assumed that unless indicated otherwise, each new keystroke
sequence starts with clear registers. Calculator registers are cleared with
the keystroke sequences
j2ndllCLR WORK
I
I
CE/C and
I
12ndllCLR TVMllcE/cl.
It will also be assumed that the calculator is operating in US date format
and US commas and decimals format, with the display showing 9
decimals. These are the default settings for the calculator, but they can be
changed in the "FORMAT" work sheet, which is accessed with the
keystroke sequence 12ndIIFORMATI. Although the number of decimals
to display is set to 9, in the examples below it will often be the case that
dollar amounts are written as rounded to the nearest .01.
2
CHAIN (CHN) AND ALGEBRAIC OPERATING (ADS) SYSTEM MODES
When the calculator is operating in chain calculation mode, the usual
algebraic order of operations is not respected. For instance, the keystroke
sequence 1+ 2 x 3 I;] results in an answer of 9. This is true because the
calculation of 1+ 2 is performed first, resulting in 3, which is then
multiplied by 3, resulting in 9. When the calculator is in AOS mode, the
result of the keystroke sequence above will be 7. This is true because in the
hierarchy of algebraic operations, multiplication is done before addition, so
2 x 3 is calculated first, resulting in 6, and then the addition operation is
applied resulting in I + 6, which is 7. The order of operations mode can be
selected in the "FORMAT" worksheet.
ACCUMULATED AND PRESENT VALUES OF A SINGLE PAYMENT
USING A COMPOUND INTEREST RATE
Accumulated values and present values of single payments using annual
(or more general periodic) effective interest rates can be determined
using the calculator functions as described below.
ACCUMULATED VALUE:
We use Example 1.1 to illustrate this function.
A deposit of 1000 made at time 0 grows at effective annual interest rate 9%.
The accumulated value at the end of 3 years is 1000(1.09)3 =1,295.03.
This can be found using the calculator in two ways.
1. We use standard arithmetic operators
with the following keystrokes.
in standard calculator
mode
1.09 [Z] 3 I;] @ 1000 I;]
The screen should display 1,295.029.In this function, y=I.09 and x=3.
2. We use time value of money functions (TVM).
12ndilPNI
ill I (this sets 1 compounding period per year).
12ndilQUITI (this returns calculator to standard-calculator mode)
3
1000 Ipvl (this sets PV to 1000),
9 Iwl (this sets the annual interest rate at 9%)
3
I.HJ
(this sets the number of years to 3),
ICPTIIFVI (this computes the accumulated value, also called future
value).
The screen should display -1,295.029.
The calculator interprets the PV of 1000 as an amount received (a cash
inflow) and the FV as the amount that must be paid back (a cash outflow),
so the FV is a "negative" cashflow. If the PV had been entered as
-1000, then FV would have been positive. This is part of the "sign
convention" used by the BA II PLUS.
PRESENT VALUE:
We use Example 1.5(a) to illustrate this function.
The present value of 1,000,000 due in 25 years at effective annual rate
.195 is 1,000,000i5 = 1,000, 000(1.195f25 = 11,635.96.
This can be found using the calculator in two ways:
1. 1.195!Z] 25 1+/-1 g GJ 1000000g
The screen should display 11,635.96. This keystroke sequence can
be replaced by:
1.195 11/x[ !Z] g GJ 1000000g
2. Using time value of money functions, we have
12ndilPNI
[I] I IENTER! 12nd! IQUIT!
1000000 IFVI19.5IWI
25 I.HJ ICPTllpvl.
The screen should display -11,635.96. (the earlier comment about the
negative value applies here).
As a more general procedure, in the equation (PV)(I+i)N =FV, if any 3
of the 4 variables PV, i, N, FV are entered, then the 41hcan be found
using the ICPTI function.
4
UNKNOWNINTEREST RATE:
As an example of solving for the interest rate, we consider Example 1.5(c).
An initial investment of 25,000 at effective annual rate of interest i grows to
1,000,000 in 25 years. Then 25,000(1+i)25 = 1,000, 000, from which we
get
i=(40)1/25 -1=.1590(15.90%).
This can be found
using
the
calculator power function with the following keystrokes:
40 I.2J .04
[;]g
1
[;], the
screen should display 0.158997234.
Using financial functions, the keystroke sequence solving for i is
12ndilPNI
ill 1 12ndilQUITI
25000 !PVll000000
1+/-IIFVI
25 [EJ ICPTI IINI
The screen should display 15.89972344 (this is the % measure).
UNKNOWN TIME PERIOD:
As an example of solving for an unknown time period, suppose that an initial
investment of 100 at monthly compound rate of interest i grows to 300 in n
months at monthly interest rate i=.75%. Then 100(1.0075t =300, from
which we get n =
Inl~~75 = 147.03
months. This can be found using the
calculator ILNI function.
Using financial functions, the keystroke sequence solving for n is
\2ndIIPN\ ill 1 12ndilQUITI
100 Ipvl 300 1+/-IIFVI
.75 IINllcPT\ [EJ.
The screen should display 147.03026. Slightly more than 147 months of
compounding will be required. The calculator returns a value of n based
on compounding including fractional periods, so that the value of
147.03026 means that 100(1.0075)14703026
= 300.
5
ACCUMULA TED AND PRESENT VALUES OF A SINGLE PAYMENT
USING A COMPOUND DISCOUNT RATE
Present and accumulated values of single payments using an effective
rate of discount can be made in the following way. Clear calculator
registers before starting the keystroke sequence.
Present Value Usin!! a Compound Discount Rate:
The present value of 500 due in S years at effective annual rate of
discount 8% is 500(1-.08)8= 500(.92)8 = 256.61.
This can be found using the calculator in a few ways:
1. We use standard arithmetic operators in standard calculator mode
with the following keystrokes.
~
.92 !ZJ 8 ~ @ 500
The screen should display 256.61.
2. 500 IFVI 8 0
[!ZY]
S 1+/-llliJIENTERljCPTllpvl
The screen should display -256.61. The calculator has calculated
PV
= -FV(1+I)-N
= -500(1-.0Sr(-8)
= -500(.92)8 = -256.61.
(Remember the sign convention for payments in and payments out.)
The following keystroke sequence could also be used.
3. 500jpvl sl+/-llwl
8 lliJICPTI!FVI
We have calculated - 500(1-.08)8
= -256.61.
6
Accumulated Value Usine: a Compound Discount Rate:
A deposit of 25 made at time 0 grows at effective annual discount rate
6%. The accumulated value at the end of 5 years is
25(1-.06)-5
= 25(.94f5
=34.06
(nearest. 0 1).
This can be found using the calculator in two ways.
1. We use standard arithmetic operators in standard calculator mode with
the following keystrokes.
.94[2] 5 1+/-1
~ ~ 25 ~
The screen should display 34.06 (rounded to nearest .01).
2. 25 Ipvl 61+/-1 lIlY
I
5 1+/-llliJlcPTIIFVI
The screen should display - 34.06 (negative sign indicating outflow). As
a third approach, we could also find the effective annual interest rate and
accumulate.
CONVERSION BETWEEN EFFECTIVE
ANNUAL AND NOMINAL RATES
The nominal annual interest rate compounded m times per year can be
found from the effective annual rate of interest and vice-versa using
calculator functions as illustrated below.
Nominal Interest Rates
A nominal annual interest rate of .24 (24%) compounded monthly is
equivalent to an effective annual rate of interest of i = .2682 (26.82%).
The relationship
i
12
= (1+ "12 )
'(12)
-1 can be used, or the equivalent rates
can be found in the following way using the ICONV function.
I
I
7
Conversion from nominal annual to effective annual interest rate:
We apply the following sequence of keystrokes.
12ndilICONVI
(NOM= appears),
24 ENTERI (the nominalrate) ill ill (CIY=appears)
I
121ENTER I (number of compounding periods),
ill ill (EFF= appears)
ICPTI.
The screen should display 26.82. We have converted the nominal annual
interest rate of 24% (keyed in as 24) compounded monthly (keyed in as
12) to the equivalent effective annual interest rate of26.82%.
Conversion from effective annual to nominal annual interest rate:
12ndilICONVI
ill
26.82IENTER!
ill (CIY= appears)
(EFF= appears)
12 IENTER! ill (NOM= appears)
ICPTI
The screen should display 23.9966 (round to 24). We have converted the
effective annual interest rate of 26.82% (key in 26.82) to the equivalent
nominal annual interest rate compounded monthly (key in 12) of24%.
Nominal Discount Rates
The nominal annual discount rate compounded m times per year can be
found from the effective annual rate of interest and vice-versa using
calculator functions as illustrated below.
A nominal annual discount rate of .09 (9%) compounded quarterly is
equivalent to an effective annual rate of interest of i = .0953 (9.53%). The
relationship
i
= (1 -
d~4)r4 -1 can be used, or the equivalent rate can be
found in the following ways.
8
Use the keystroke sequence 12ndilPNI 4 IENTERI12ndIIQUITI. This
sets the number of compounding periods per year to 4 and returns to
standard calculation functions. It is important to do this step first,
entering the number of compounding periods in the year.
I. We first find the equivalent effective annual rate of interest from the
given nominal annual rate of discount.
[fY]
41+/-1 [EJ 91+/-llwll
ICPTIIFVI
The display should read -1.0953; we interpret this as indicating that the
effective annual rate of interest is 9.53%. We have calculated
p
( 1+
1
-N
-.09
1+4
) =(
-4
)
= FV = -1.0953,
where P = 4 was entered with the PN function, N = -4 was entered with
41+/-1 [EJ,and 1=-.09
(or d=9%) wasenteredwith91+/-llwl.
2. We now find the equivalent nominal annual rate of discount from the
given effective annual rate of interest.
41+/-1 [EJ, Key in 1.0953ll+/-IIFVI,
Key in llpvl, Key inICPTIIINI.
The display should read -9.00; this is the negative of the equivalent
nominal annual rate of discount compounded 4 times per year.
Note that when we enter FV, we enter - (l+EFF).
LEVEL PAYMENT ANNUITY VALVAnON
The accumulated value and present value of a level payment annuityimmediate can be found using calculator functions. Clear calculator
registers before starting the keystroke sequence. The calculator should be
in standard calculator mode.
9
Accumulated
Value of Annuitv-Immediate:
A deposit of 1000 is made at the end of each year for 20 years. The
deposits earn interest at an effective annual rate of interest of 4%. The
accumulated value of the deposits at the time of (and including) the 20th
deposit
is
1000s2O].o4
= 1000[ (1.~~:O-l] = 29,778. This can be found
using the calculator.
Use the keystroke sequence 12ndllPNIl
IENTERI12nd[
I
QUIT! (this
sets the number of compounding periods per year to 1 and returns to
standard calculation functions). It is important to do this step first, so that
the compounding period corresponds to the payment period.
20 IRJ41INI
0 !PVllOOO IPMT!lcPT!IFV!
The display should read -29,778.08, the negative of the accumulated
annuity value.
Present Value of Annuity-Immediate:
Payments of SO will be made at the end of each month for 10 years. The
monthly compound interest rate is % %. The present value of the annuity
one month before the first payment is made is
SO~.OO75 =SO[1;~~5 J=3,947.08
(10 years, 12 months per year).
The calculator should still have /2nd IIPN set to 1 (if not, use the
keystroke sequence outlined above to set PN to 1). This means that the
rate entered as IN will be a rate per period. In this example the period is
one month. We use the following sequence of keystrokes.
I
120 lli] .7S!INIIENTERI
50 IPMTIIENTERI
0 IFVllcPTllpvl
The display should read -3,947.08, the negative of the present value of
the annuity.
10
In the general equation for an annuity-immediate
PV
= PMT.
ami if any 3
of the 4 variables PV, PMT, N, i (in %) are given, then the calculator
functions can be used to solve for the 4th variable. The same is true for the
equation PMT . SNji = FV (keep in mind that when PMT is entered as
positive, FV or PV are returned as negative, and vice-versa).
Findin2 the Payment Amount:
A loan of 1000 is to be repaid with monthly payments for 3 years at a
compound monthly interest rate of .5%. The monthly payment is K
where
1000 = Ka361
. 005'
so that K = alOOO = 30.42.
'361.005
This can be found using the following sequence of keystrokes:
36 lliJ .5 IINIIENTERI
1000 Ipvl 0 IFVllcPTllpMTI
The display should read -30.42.
Findin2 the Interest Rate:
Suppose that the loan payment is 35 for a 36 payment loan of amount
1000 and the interest rate is to be found. Then 1000 = 35ll36Ji' There is no
algebraic solution for i. The following keystrokes give us i.
36lliJ 1000 1+/-1 Ipvl 35 IPMTI 0 IFVllcPTllwl
The display should read 1.31(%). That is the effective rate of interest per
month.
Findin2 the Number ofPavments:
We will use Example 2.13 to illustrate the calculator function for finding
the unknown number of payments. In Example 2.13, Smith wishes to
accumulate 1000 by means of semiannual deposits earning interest at
nominal annual rate i(2) = .08, with interestcreditedsemiannually.
In part (a) of Example 2.13, Smith makes deposits of 50 every six
months. We wish to solve for n in the equation 1000 = 50 . snJ.04.
11
The following keystrokes give us n.
1000 IFvl 50 1+/-llpMTI
41INI
0 IpvllcPTI
[R]
The display should read 14.9866. 14 deposits are not sufficient and 15
full deposits are more than sufficient. The accumulated value of the
deposits just after the 14thdeposit is 50,si4L04=914.60, so an additional
deposit of 85.40 is needed at the time of the 14thdeposit to bring the total
accumulated value to 1000. The accumulated value 6 months after the
14th deposit
is
50. Si4l04
= 50(1.04)
. si4L04 = 951.18 ,
so a deposit of 48.18 is needed at time 15 to bring the accumulated value
to 1000.
The next functions reviewed relate to finding the value of an annuity-due.
Annuitv-Due:
The accumulated value and present value of a level payment annuity-due
can be found using calculator functions. The same methods apply that were
used for annuities-immediate, with the additional requirement that
keystrokes 12nd!IBGNI12nd!ISETI!2ndIIQUITI must be entered to
make the calculator view payments as being made at the beginning of each
period (there is a screen indication of the BGN mode when it is invoked;
when in BGN mode, in order to return to END mode, use the keystrokes
!2ndIIBGNI12ndIISETI12ndIIQUITI, and BGN should disappear from
the screen display, and the calculator is in END mode). In the equation
PV = PMT. i:i:vJi'if any 3 of PV, PMT, N, i are entered, we can find the
4th.In the equation FV = PMT. sm; any 3 of FV, PMT, N, i are entered,
we can fmd the 4th(as before, PMT and PV or FV are opposite signs).
Findine the Time and Amount of a Balloon Payment:
We can use the calculator functions to find the balloon payment required
to repay a loan which has level payments for as long as necessary with a
final balloon payment. In Example 2.15(a) of Chapter 2, a loan of5000 is
being repaid by monthly payments of 100 each, starting one month after
the loan is made, for as long as necessary plus an additional &actional
12
payment at the time of the final regular payment. At interest rate
;(12)
= .09, we are to find the number of full paymentsthat are required
to repay the loan, and the amount of the additional fractional payment
required if the additional fractional payment is made at the time of the
final regular payment. We find the number of payments needed with the
following keystrokes (the calculator should be in END mode).
5000 !PVllOO 1+/-llpMTI .75 IIIYllcPTllliJ
The display should read 62.9. This indicates that the 62ndpayment is not
quite enough to repay the loan. The additional payment needed, say X, at
the time of the 62ndregular payment of 100 is found from the relationship
X = 5000(1.0075)62 -100,s621.0075 = 89.55. The keystrokes that will
produce the value of X are
5000 IpvllOO 1+/-llpMTI .75 IINI 62lliJlcPTIIFVI
The display should read -89.55.
ANNUITIES WHOSE INTEREST AND PAYMENT PERIODS DIFFER
A function is available to calculate annuity values when the interest
period and payment period do not coincide. The following examples
illustrate the use of this calculator function.
I. Annuity-immediate of 10 annual payments of 1 each, with interest at
a nominal annual rate of 8% compounded quarterly. The present
value one year before the first payment is aiO!j' where
j = (1.02)4-I = .082432
is the effective annual rate of interest that is equivalent to the nominal
annual. The present value will be awJ.082432= 6.6367. The annuity
value can be found using financial functions with the following
sequence of keystrokes:
12nd! [lli] lIENTER! (this sets 1 payment per year),
ill 4 ENTER (this sets CN=4 interest conversion periods per
I
I
year), 12ndIIQUITI.
13
10 lliJ (10 annual payments),
8 I IIY (nominalannual interestrate of 8%),
I 1+/-llpMTllcPTllpvl.
I
The display should read 6.637, the present value of the annuity. To
find the accumulated value of the annuity, we continue with the
followingkeystrokes: 0 Ipvl!cPTllFvl
The display should read 14.655, the accumulated value of the annuity.
Note that if we find the equivalent effective annual rate of interest first,
i = .082432 , we could have found the annuity value as follows without
setting CN to 4. This is done in the following way (CN and PN are
both set to I).
10 lliJ (10 annual payments),
8.2432 lIlY (effective annual interest rate),
I
I 1+/-llpMTllcPTllpvl
The display should read 6.637, the present value of the annuity.
2. Annuity-immediate of I per month for 5 years at effective annual
rate of interest 6%. The equivalent one-month interest rate is
j = (1.06)1112 -I = .004868 and the present value of the annuity
one month before the first payment is a60lj = 51.924. The annuity
value can be found using financial functions with the following
sequence of keystrokes.
/2ndl [fZY] 12 ENTER (this sets 12 payments per year),
I
I
ENTER (this sets CN=I interest conversion period per year),
then 12ndIIQUlTI.
ill 1
I
I
5 j2ndllxPNIlliJ
(5 x 12 = 60 monthly payments),
6 lIlY I (effective annual interest rate),
I 1+/-llpMTllcPTllpvl
The display should read 51.924, the present value of the annuity.
14
As with Case 1 above, we could have used an alternative approach based
on the one-month effective interest rate of .004868. PN and CN should
be set to 1 before the following keystrokes.
60 lliJ .4868 IINI (monthly effective interest rate),
1 1+/-llpMTllcPTllpvl
The display should read 51.923, the present value of the annuity.
VALUATION OF INCREASING AND DECREASING ANNUITIES
The values of (Ia}ii'li (present value) and (Ds}iili (accumulated
value)
can be found using calculator financial functions. From those values we
can then find (Is)nri = (Ia)nri . (1 + it, and (Da)nri = (Ds)nri . vn.
The following two examples illustrate the method.
Finding (Ia)1i1i :
..
Suppose that we wish to find (Ia)Wl.os
=
20 v20
amos.
.08
. We first find the
numerator with the following keystrokes.
12ndllBGNI12ndiiSETI12ndliQUITI
20 lliJ 8 IINII
1+/-llpMTI
20 IFVllcPTllpvl
The display should read 6.3126, which is aWl.os - 20v20 .
In this sequence of keystrokes, we have created a series of 20 payments
received of 1 each at the start of each year
(IBGN I), combined
with a
payment of 20 paid out at the end of 20 years dFVI). The net present
value is aWl.os - 20v20
= 6.3126.
Then, (Ia )Wl.os= 6'~01826- 78.908.
Note that instead of entering 1 1+/-llpMTI we could enter
1g
.08 B 1+/-llpMTI, and instead of entering 20lFVI we could enter
20 g .08 B IFvl. Then ICPTllpv! includes division by .08. Note that
15
we cannot use these calculator functions to find (Is)-,.,
but since the
nil
numerator of (Is)-,.
is s-,.
- n, we can find s-,.
first, then subtract n,
nil
nil
nil
and then divide by i.
Finding (Ds)nJi :
35(1.04i5 -sm04
. We first
.04
find the numerator with the following keystrokes. The calculator
payment mode should be set to END.
Suppose that we wish to find (Ds)m04
35 lli] 41wliENTERI
1 IPMTI ICPTIIFVI
=
35 1+/-llpvl
The display should read 64.4609, which is 35(1.04)35 - sm04.
In this sequence of keystrokes we have created an initial payment received
of 35 at time 0 ( PV ), and a series of 35 payments of 1 each paid out at the
I
I
end of each year. The net accumulated value at the end of 35 years is
35(1.04)35 -s351.04
= 64.4609
(lFVI).
Then,
(Ds)35104=
64.4609
= 1,611.52.
.04
Note that we could have incorporated division by .04 into the keystroke
sequence by keying in 35
g
.04
B 1+/-llpvl insteadof
351+/-llpvl, and by keying in 1g .04 B IPMTI insteadof 1 ]PMTI.
It is also possible to use the cashflow worksheet to find present and
accumulated values of a series of non-level payments. There is a
limitation that allows an initial payment, CFo, and up to 24 more
payment amounts.
This would be a less efficient way of finding (Ia )20108' for instance.
16
DEPRECIA nON
An asset has a purchase price of 1200 and a salvage value of 100.
According to the sum-of-years-digits method with a 10-year depreciation
period, the depreciation amounts and book values will be
D)
= 1~ .(1200-100)=200,
D2 =
is '(1200-100)
= 180,...,DIO = 20,
BJ = 1000, B2 = 820,..., B9 = 120, BIO= 100.
These values can be found using the calculator depreciation worksheet
with the following sequence of keystrokes.
1 [2nd! IDEPR[, then 12ndliSETI until SYD shows in the display.
Then,
m
gets us to LIF=, and we key in IOIENTERI
(depreciation
Then
m
over 10 years).
gets us to MOl =, which we skip,
and [IJagain gets us to CST=, and we enter 1200IENTERI (initial cost).
Then
m
gets us to SAL=, and we enter 100 ENTER (salvage value).
Then
m
gets us to YR=, and we can enter any year, say SIENTERI.
I
I
Then [I] gets us DEP=120, which is Ds;
Then
m
gets us RBV=400, which is Bs.
We can use the same functions for declining balance (a discount rate
must be entered), and straight line depreciation.
17
LOAN AMORTIZATION
For a loan with level payments, there are calculator functions for finding
outstanding balances, interest or principal paid in a single payment, and
interest or principal paid in a range of payments. Parameters are entered
for PN, N, IN, PV as with the usual annuity valuation. The calculator
will give us the payment using the PMT key. The following example
illustrates how we get the amortization quantities.
We use Example 3.3 to illustrate these functions. A homebuyer borrows
$250,000 to be repaid over a 30-year period with level monthly payments
beginning one month after the loan is made. The interest rate on the loan
is a nominal annual rate of 9% compounded monthly. The loan payment
is K =
250,000
= 2,011.56 .
~OO75
The outstanding balance at the end of the first year (after the Ith monthly
payment) is OB12 = 2, 0 11.56~.oo75 =248,292.0 I. The principal repaid
in the Ith payment is PR12 = Kv360-12+1 = 2,011.56v349= 148.25,and
the interest paid in the 12thpayment is 112=K(l-v360-12+1 )=1, 863.30. The
principal repaid in the 2ndyear (the 13ththrough 24th payments inclusive) is
PR 13 + PR 14 +". + PR 23 + PR24
348
= K( V
+v
347
+" .+v
338
+v
337 )
= 1,868.21,
the interest paid in the 2nd year is
I 13 + I 14 +".+
I 23
+
I 24 = K(l -v 348 + I -v 347 +".+l-v
=12K -(PR13+PRI4+.,,+PR23+PR24)
338
+l-v
337
)
= 22,270.46.
These calculations can be completed using calculator functions. We first
clear the calculator registers.
12ndlIPNII
IENTER
I !2ndIIQUITI
360 lli1 .75 Iwl250000
(sets payment period to 1)
IpvllcPTllpMTI
18
The value -2,011.56 is returned (we must key in ICPTllpMT! in order
to use the amortization functions).
To find OB12 use the following keystrokes.
12nd[IAMORTI
(this opens the amortization worksheet).
The display reads PI =. Key in 12 IENTER! ill.
The display reads P2 =. Key in 12IENTER[.
Then using
ill
again, the display should read BAL = 248, 292.01.
This is OB12'
Using
ill
again gives the display PRN= -148.25;this
Using
ill
again gives the display INT= -1,863.30; this is -112'
is -PR12'
To find 113+114+...+123+124
and 1i3+1i4+",+P23+P24
use the
following keystrokes while still in the amortization worksheet (we leave
a worksheet using the 12nd!IQUlTI sequence of keystrokes).
Use
ill
until PI= appears again and enter 13 (key in 13 ENTER ).
Then use
I
ill and
I
P2= appears and we enter 24 (key in 24 [ENTER [).
The next use of ill given us BAL = 246, 473.79 , the outstanding balance
at the end of the period to time 24 months).
ill
again gives us PRN = -1,868,21;
this is -(1i3 + 1i4 + . .. + P23 + P24) ,
the negative total amount of principal paid in payments 13 to 24 (the
second year).
ill
this is -(113 + 114+... + 123+ 124)'
again gives us INT= -22,270.46;
the negative total amount of interest paid in payments 13 to 24.
Note that 1i3 + 1i4 +... + P23 + P24
=
OB12
- OB24 could be found from
OB12 and OB24, and
113+114 +...+123
+124
= 12K -(PR13+PRI4 + ... +PR23+PR24)
= 12K
-(OB12-0B24)'
19
BOND VALVA TION AND AMORTIZATION
It is possible to calculate the price or yield to maturity of a bond using
the calculator's bond worksheet. We use Example 4.1(a) to illustrate the
bond worksheet functions.
Findin!! The Bond Price on a Coupon Date:
A 10% bond with semiannual coupons has a face amount (par value) of
100 and is issued on June 18, 1990. The bond has a maturity date of June
18, 2010. We wish to find the price of the bond on its issue date using a
nominal annual yield rate of 5% convertible semi-annually.
= 162.76
The bond price is 100v6~5 +100(.05)'~.025
(nearest .01). If
the bond has maturity value 110. then the price is
40
110v025 +100(.OS)'~.025
= 166.48.
These prices can be found using the following sequence of keystrokes.
12ndilBONDI
The display
(this opens the bond worksheet).
shows SDT = (there may be a date here).
Key in 6.1890 ENTER (valuation date, June 18, 1990)
I
I
Using ill gives us CPN=,and we key in 10 ENTER
I
I
(coupon amount of 10 per year).
Using ill gives us RDT=, and we key in 6.1810 I ENTER
I
(redemption date 20 years later, June 18,2010).
Using ill gives us RV=, and we key in 100 IENTERI.
(redemption amount of 100)
Using ill gives us ACT, which we leave
(refers to actual day count for bond valuation).
Using
ill
gives us 2/Y (we leave this setting since the coupons are paid
twice per year; for coupons once per year we would usel2ndiiSETI to
change the display to 1/Y, which means we want "1 coupon per year").
Using ill gives us YLD=, and we key in SIENTERI (nominal yield rate
per year, compounded 2/Y).
20
Using
ill gives us PRI=, and using ICPTlresults in 162.76 (nearest .01)
on the display. This is the bond price based on a face amount of 100.
Multiplying by 1,000,000 gives the price for a 100,000,000 face amount
bond.
To change the maturity value to 110, we use ill until we reach RV=, and
we enter 110. Then use ill until we get to PRI= again, and use ICPTI.
The resulting display should be 166.48.
Findine The Bond Yield on a Coupon Date:
We can use the worksheet to find the yield rate from the price. Suppose
that the bond above with face amount 100 has a price of 150. There is no
j(2) 2}, where}, the 6algebraic solution for the yield-to-maturity
=
month yield rate is the solution of the equation 150 = 1OOv~o + 5a:wJj'
The yield-to-maturity
worksheet as follows.
can be found using the calculator
in the bond
SDT=6-18-1990 (enter 6.1890),
CPN=10, RDT=6-18-2010 (enter 6.1810),
RV=100, ACT, 2N, bypass YLD=, PRI=150.
Use ill to get to YLD= and then use ICPTI to calculate the yield rate.
;(2)
The display should read 5.76 (YTM is
=.0576, which is) = 2.88%
every 6-months).
The previous calculations can also be done using the calculator annuity
functions as follows.
12nd[ IPNI 2 IENTERI12nd[ !QUITI
(ensures yield period and coupon period are the same).
40 lliJ (40 6-month periods),
5 Iwl (nominal annual yield rate),
5 1+/-llpMTI
(coupon payment every 6 months),
100 I+/-IIFV[
(maturity value),
ICPTllpv[ should result in 162.76.
21
For the bond with maturity amount 110 we do the following.
40 lliJ 5 IllY I 5 1+/-llpMTIII0
]+/-IIFVllcPTllpvl
This should result in 166.48.
To find the yield rate for a price of 150 we do the following.
40 lliJ 150 Ipvl 5 1+/-llpMTII00
1+/-IIFVllcPTIIIIYI
This should result in 5.76 (the nominal annual yield).
Bond Amortization:
The bond amortization components can be found using the calculator's
amortization worksheet in much the same way they are found for loan
amortization.
A bond has face amount 1000, coupon rate 5% per coupon period,
maturity value 1000, 20 coupon periods until maturity and yield-tomaturity 6% (per coupon period). The bond's amortized value just after
the 5thcoupon is
BVs
= 1000vI~6
+ 1000(.05).
a151.o6
= 902.88.
This can be found using the following keystrokes.
12ndllPIYll
IENTERIIZndIIQUIT!
ZO lliJ 6 IIIYI 50 1+/-llpMTIIOOO
1+/-IIFVllcPTllpvl
The 20-year bond price of 885.30 should appear.
Then IZndllAMORTI should result in PI =. Key in 5]ENTERI ill.
This should result in P2 =, and again enter 5 jENTERI.
Then ill should result in 902.88, the balance just after the 5th coupon.
Using the ill key again gives PRN=3.94, which is the negative of the
principal repaid in the 5thpayment (the amount of write-up is 3.94, since
the bond was purchased at a discount). Using the ill key again gives
INT= - 53.94, which is the negative of the amount of interest due in the
5th payment.
22
Bond Price and Yield Between CouDon Dates:
The bond examples considered so far have had valuation take place on a
coupon date. It is also possible to use the bond worksheet functions to
find the price (given the yield) or the yield (given the price) of a bond at
any time, on or between coupon dates.
We use Example 4.2 to illustrate the valuation of a bond between coupon
dates. A bond has face amount 100, with an annual coupon rate of 10%
and coupons payable semi-annually. The bond matures on June 18,2010
and is purchased on August 1,2000 at a yield rate of 5% (nominal annual
yield compounded semi-annually). The quoted purchase price from
Example 4.2 is 138.60. This can be found using the following keystrokes
in the bond worksheet.
SDT=8-01-2000 (enter 8.0100),
CPN=10, RDT=6-18-2010 (enter 6.1810),
RV=100, ACT, 2N, YLD=5 dENTERI must be used after each entry).
At PR1=, use ICPTI to calculate the price.
The display should read 138.60. Note that this is the quoted price which
excludes the accrued coupon. The accrued coupon amount is found at
AI=.
If a price had been entered instead of a yield rate, we could have
computed the yield.
INTERNAL RATE OF RETURN AND NET PRESENT VALUE
The internal rate of return for a series of payments received and
payments made can be found in a couple of different ways, depending
upon the nature of the series of payments. When we consider a level
payment annuity with or without a balloon payment at the time of the last
annuity payment, we can enter values into the variables
lliJ, Ipvl, IPMTI and IFvl, and then use !cPTI Iwl to find the
interest rate which satisfies the relationship PV
The internalrate of return is).
= PM!'.
a;;/j+ FV . vi.
23
We can use the cashflow worksheet (lCFI) to enter a cashflow at time 0,
CFo, along with up to 24 additional cashflows at the end of 24
successive periods, CO1,CO2,..., C24. Once these cashflow amounts are
entered, we can use the IIRRI function (internal rate of return) to
calculate an internal rate of return. It is a solution) to the relationship
CFo +C01'vj
+C02.v;
+...+C24.v;4
= O.
The cashflow amounts can each be positive (an amount received) or
negative (an amount paid out).
Calculatine Internal Rate of Return
We illustrate how the internal rate of return in Example 5.1 can be found
in this way. Example 5.1 has the following series of cashflows, where
time is measured in 6-month intervals:
Co = - 5100, C1 = 0, C2 = -2295, and C3 = 7982.5 .
The cashflow worksheet is cleared using ICFI12ndllCLR
WORKI
The following series of keystrokes solves for the internal rate of return),
where) will be the 6-month internal rate of return:
Key inlCFI, the display should read CFo=,
Key in 5100
0
I
ENTER III, the display should read CO1=,
I
Key in 0 ENTER III III, the display should read C02=,
I
I
Key in 2295 0
IENTER I III III, the display should read C03=,
Key in 7982.5 ENTER ,
I
I
Key in IIRRllcPTI. The display should read 3.246.
This is the 6-month internal rate of return.
As seen in Example 5.2, a series of cashflows may have a unique internal
rate of return, it may have more than one internal rate of return, or it may
have no internal rate of return. If we attempt to solve for the internal rate
of return with the calculator function for Example 5.2, we get the
following results:
24
5.2(a) Co = -1, C] = 2.3, and C2 = -1.33, and the calculator returns an
"Error 7" message when IIRRI ICPTI is keyed in, which generally
occurs when there is no solution for the internal rate of return.
5.2(b) Co =-I,C] =2.3, and C2 =-1.32, and the calculator returns an
IRR of 10%. This is the smaller of the two solutions, 10% and 20%.
5 .2(c) Co = -1, C] = 2.3, C2 = -1.3125, and the calculator returns an IRR
of 5%, again the smaller of the two solutions, 5% and 25%.
5.2(d) Co = -1, C1 = 2.3, C2 = -1.2825,
and the calculator returns an IRR
of - 5%, again the smaller of the two solutions, - 5% and 35%.
Calculatine: Net Present Value:
The calculator INPvl function can be used to calculate the present value
of a series of cashflows that have been entered into the cashflow
worksheet. We use Example 5.2(a) as an illustration. Suppose that we
wish to find the present value at time 0 of the three cashflows
Co = -1, C] = 2.3, and C2 = -1.33,
using an interest rate of 10%.
This would be -1 + 2.3v] -1.33vl. We can find this present value as
follows. We enter the cashflows Co = -1, C] = 2.3, and C2 = -1.33, into
the ICFI worksheet as shown above. Immediately after C2 has been
entered, and while still in the cashflow worksheet, press INPvl. The
display will read 1=0. We key in 10 ENTER rn ICPTI.
I
I
The display should read NPV = - 0.008264. This is the present value of
the cashflows valued at an interest rate of 10% per period.
In the cashflow worksheet it is possible to specify an initial cashflow
amount CFO, and 24 additional amounts C I to C24. Each cashflow
amount may be made multiple times ("grouped" cashflows of the same
amount made at successive points in time).
25
We use Example 2.5 to illustrate this application. Suppose that 10
monthly payments of 50 each are followed by 14 monthly payments of
75 each. If interest is at a monthly effective rate of I%, what is the
accumulated value of the series at the time of the final payment? The
following keystrokes give us the accumulated value.
[Q] 0
I
ENTER
I
Key in 50 ENTER
I
[I]
I
[I] 10 ENTER
I
I
[IJ (this sets a payment amount of
C01=50 to be made for F01=10 successive periods),
Key in 75 ENTER
I
I
[I] 14 jENTER
I [I]
(this sets a payment amount of
C02=75 to be made for F02=14 successive periods after the first 10
periods).
Key in INPvll
jENTERI [IJ ICPTI
The screen should display 1,356.47. This is the present value of the 24
payments one month before the first payment.
With NPV = 1,356.47 still displayed, Key in Ipvl.
Key in 12ndilQUITI 24/R] I !INllcPTIIFVI
This should result in the display FV = -1,722.36. This is the accumulated
value of the series of24 payments at time of the final payment.
As a variation on Example 2.5, suppose that 10 deposits of 50 per month
are made into an account earning monthly interest rate i. Suppose further
that one month after the lOth deposit, monthly withdrawals are made
from the account of amount 75 per month. The account balance is 0 just
after the 14th withdrawal. We wish to find the monthly interest rate on the
account. We wish to solve for i in the equation 50siOli = 75aI4li'
This equation is equivalent to 50aiOli
= 75vlOaI4li'
We can place this in
the context of internal rate of return, where we wish to find the internal
rate of return for a sequence of cashflows of 10 payments paid of 50 each
(C01=50, F01=10) followed by 14 payments received of 75 each
(C02=-75, F02=14), and whose total initial present value is CFO=O.
26
We use the following keystrokes:
[g]
0 ENTER
I
I []
50 ENTER
I
I []
10 ENTER
I
751+/-1 I ENTER ill 14IENTERIIIRRllcPTI.
1
The display should read IRR = 6.518.
I
[]
27
EXAMPLES FROM SOAJCAS EXAM FM/2
(FORMERLYCOURSE2 COMPOUNDINTEREST)
2003, #33 (Annuity Valuation)
L::.
8
At an effective annual interest rate of i, i > 0" both of the following
annuities have a present value of X:
(i) A 20-year annuity-immediate with annual payments of 55
(ii) A 30-year annuity-immediate with annual payments that pays 30 per
year for the first 10 years, 60 per year for the second 10 years, and 90
per year for the final 10 years.
Calculate X
(A) 575
I
(B) 585
SOLUTION
(C) 595
(D) 605
(E) 615
I
The series of cashflows representing the difference between (i) and (ii) is
a series of 10 payments of 55 - 30 = 25 each, followed by a series of 10
payments of 55 - 60 = - 5 each, followed by a series of 10 payments of
-90 each. The interest rate that makes the present value of this series
equal to 0 is found using the IRR function as follows.
Key in
[g]
0 IENTER! ill 25 IENTER! ill 10 ENTER ,
I
ill 5 I+/-IIENTERI
I
ill 10 IENTER\,
ill 90 1+/-1 ENTER ill 10 IENTERI,
I
IIRRllcPTI.
I
The display should read IRR = 7.177.
This is the interest rate per year. Set PlY and CIY to 1. The following
keystrokes give us the value of X
12ndilQUITI12ndllCLR
20 lliJ 7.177!wl
TVMI
551+/-llpMTllcPTllpvl.
The display should read 574.74. Answer: A
28
2003, #8 (Annuity Valuation)J
(This is also Exercise 2.2.2)
l.::t
Kathryn deposits 100 into an account at the beginning of each 4-year
period for 40 years. The account credits interest at an effective annual
interest rate of i. The accumulated amount in the account at the end of 40
years is.x; which is 5 times the accumulated amount in the account at the
end of 20 years. Calculate X
(A) 4695
I
(B) 5070
SOLUTION
(C) 5445
(D) 5820
(E) 6195
I
We denote the 4-year rate of interest byj. Then the accumulated value at
the end of 40 years is X = 100siOlr (10 4-year periods, with valuation
one full 4-year period after the 10th deposit). The accumulated value at
the end of 20 years is 100s51r
Weare
given that 100SiOlj
= 5 x 1OOs51j' which
100SiOlj - 5 x 100s51j
is the same as
=0
(after we multiply both sides of the equations by Vj)' This can be
interpreted as saying that 5 payments of 100 per period, followed by 5
payments of -400 per period has an accumulated value (and present
value) ofO. We can use the IRR function to find} as follows.
Key in
ill
[g] 0
I
ENTER ill 100 ENTER ill 5 ENTER ,
I
400 1+/-1 ENTER I
I
I
I
ill5IENTERIIIRRllcPTI.
The display should read IRR = 31.95.
The 4-year interest rate is } = .3195.
Then we apply the following keystrokes.
12nd[IQUITI12ndIICLR
TVMI
12nd[IBGNI12ndIISETI12ndIIQUIT/
10 lliJ 31.95 IIN[
100 IPMTllcPTI
IFV[.
The display should read -6,194.44.
Answer: E
I
I
29
2003, #26 (Decreasing AnnUity~
'-='
1000 is deposited
(Exercise 2.3.13)
into Fund X; which earns an effective annual rate of
6%. At the end of each year, the interest earned plus an additional 100 is
withdrawn from the fund. At the end of the tenth year, the fund is
depleted. The annual withdrawals of interest and principal are deposited
into Fund Y, which earns an effective annual rate of 9%. Determine the
accumulated value of Fund Yat the end of year 10.
(A) 1519
(B) 1819
SOLUTION
I
(C) 2085
(D) 2273
(E) 2431
I
The 10 deposits to Fund Yare 160, 154, 148,..., 112, 106.
These can be entered as 10 separate cashflows in the ICFI worksheet.
The NPV function at 9% will give a present value (one year before the
deposit of 160) of 880.59. Then the FV at the end of 10 years will be
2,085. Answer: C
November2001,#16(Increasinf AnnUi~~
(Exercise2.3.12)
~
Olga buys a 5-year increasing annuity for X Olga will receive 2 at the
end of the first month, 4 at the end of the second month, and for each
month thereafter the payment increases by 2. The nominal interest rate is
9% convertible quarterly. Calculate X
(A) 2680
I
(B) 2730
SOLUTION
(C) 2780
(D) 2830
(E) 2880
I
With monthly rate}, X = 2(Ia)601J'
We are given 3-month rate .0225, so that (1+})3 = 1.0225, and therefore,
) = .007444. The numerator of (Ia)601i can be found by the following
keystrokes.
12ndllBGNI[MJ ~
12ndllPNI12
IENTER
12ndilQUITI
I
[I] ICNI 4 IENTERI12ndIIQUITI,
60 [H] I 1+/-llpMTI 9jINI60 IFVllcPTI!pvl
This results in the display reading PV = 10.1587.
Then
g
.007444
~ 2 results in 2,729 on the display. Answer: B
30
~
2003, #25 (Loan Re~
John borrows 1000 for 10 years at an effective annual interest rate of
10%. He can repay this loan using the amortization method with
payments of P at the end of each year. Instead, John repays the 1000
using a sinking fund that pays an effective annual rate of 14%. The
deposits to the sinking fund are equal to P minus the interest on the loan
and are made at the end of each year for 10 years.
Determine the balance in the sinking fund immediately after repayment
of the loan.
(A) 213
I
SOLUTION
(B) 218
(C) 223
(D) 230
(E) 237
I
P is found as follows. Clear all registers and be sure that IN and PN are
both set to 1 and payment mode is END. Apply the following keystrokes.
10 lliJ 10 IINI1000
IpvllcPT!lpMT!
The screen displays -162.75. The deposits to the sinking fund are 62.75
each. The accumulated value of the sinking fund deposits at 14% is
found as follows.
10 lliJ 141INI 62.75 IPMTI O!PVI!CPTIIFVI
The display reads -1,213.42.
After the principal loan amount of 1,000 is paid, 213 is left in the sinking
fund. Answer: A
31
2003, #42 (Bond Amortization)
,
l.=r
A 10,000 par value 10-year bond with 8% annual coupons is bought at a
premium to yield an effective annual rate of 6%.
Calculate the interest portion of the 7th coupon.
(A) 632
I
SOLUTION
(B) 642
(D) 660
(C) 651
(E) 667
I
The AMORT functions can be used as follows.
10 lliJ 6 IINI 800 1+/-1 10000 1+/-IIFVllcPTllpvl
The display reads 11,472.02 (the bond price). Then
12ndiiAMORTI
7 IENTER [
ill 7 ENTER ill
I
I
shows BAL = 10,534.60 (the book value after the ih coupon).
Then ill shows PRN = - 158.42 (the negative of the principal repaid),
and ill shows INT = - 641.58 is the negative of the interest paid.
Answer: B